Stability of faces in asymmetric evolutionary games

Abstract

The concept of a face of population states arises naturally in evolutionary games. This paper studies faces of profiles in asymmetric evolutionary games with infinite strategy space. The concepts of strong immovable and immutable faces of profiles are introduced and stability results for these faces are discussed.

Appendix

This section begins with the proof of Lemma 3 which is stated in Sect. 3.1. The remainder of the Appendix is devoted to the proofs of various results employed in part (b) of Theorem 1. In particular, the uniform continuity of the map $t \mapsto \sigma _i(a_i|\mu (t))$ is established in Lemma 4. This is followed by Lemma 5 where we prove that $J_i(\eta _i(\cdot |{\bar{B}}), \eta _{-i}) - J_i(\eta _i,\eta _{-i}) = 0$ for any weak cluster point $\eta$ of replicator trajectories originating nearby the face $\bigtriangleup _{{\bar{B}}}$. The weak lower semicontinuity of $\Vert \cdot \Vert _{\infty }$ is verified in Lemma 6 which in turn is used to show the weak closedness of the $\varepsilon$-ball around the face $\bigtriangleup _{{\bar{B}}}$ in Lemma 7.

Proof of Lemma 3

By the definition of strong norm [(see e.g., (1) in Mendoza-Palacios and Hernández-Lerma (2015) or p. 360 of Shiryaev (1995)],

$$\begin{aligned} \Vert \mu _i - \mu _i(\cdot |{\bar{B}})\Vert = 2 \; \text {sup}_{B_i \in {\mathcal {B}}(A_i)} |\mu _i(B_i) - \mu _i(B_i|{\bar{B}})|. \end{aligned}$$

For a Borel set $B_i$, we consider two cases, namely $B_i \cap {\bar{B}}_i = \emptyset$ and $B_i \cap {\bar{B}}_i \ne \emptyset$.

Case 1: $B_i \cap {\bar{B}}_i = \emptyset$.

In this case,

$$\begin{aligned} |\mu _i(B_i) - \mu _i(B_i|{\bar{B}})|&= \left| \mu _i(B_i) \right| \; (\text {by} \; (5))\\ ~&\le \; |\mu _i({\bar{B}}_i^c)| = 1- \mu _i({\bar{B}}_i). \end{aligned}$$

Note that the maximum value $(1- \mu _i({\bar{B}}_i))$ in this case is achieved when $B_i = {\bar{B}}_i^c$.

Case 2: $B_i \cap {\bar{B}}_i \ne \emptyset$. If $B_i^1 = B_i \cap {\bar{B}}_i$ and $B_i^2 = B_i \cap {\bar{B}}_i^c$, then

$$\begin{aligned} |\mu _i(B_i) - \mu _i(B_i|{\bar{B}})|&= \left| \mu _i(B_i^1)+\mu _i(B_i^2) - \frac{\mu _i(B_i^1)}{\mu _i({\bar{B}}_i)} \right| \; (\text {by} \; (5))\\ ~&\le |\mu _i(B_i^2)| \quad \left( \because \mu _i(B_i^1) - \frac{\mu _i(B_i^1)}{\mu _i({\bar{B}}_i)} \le 0\right) \\ ~&\le |\mu _i({\bar{B}}_i^c)| = 1- \mu _i({\bar{B}}_i). \end{aligned}$$

Combining Case 1 and Case 2, we obtain

$$\begin{aligned} \Vert \mu _i - \mu _i(\cdot |{\bar{B}})\Vert = 2 \; (1- \mu _i({\bar{B}}_i)). \end{aligned}$$

$\square$

Lemma 4

Under the conditions and notations of part (b) in Theorem 1, the map $t \mapsto \sigma _i(a_i|\mu (t))$ is uniformly continuous.

Proof

Since $U_i(\cdot )$ is bounded, by Proposition 4.4 of Mendoza-Palacios and Hernández-Lerma (2015), $\sigma _i(\cdot |\mu )$ satisfies conditions (i) and (ii) of Theorem 4.3 of Mendoza-Palacios and Hernández-Lerma (2015).

As we are dealing with probability measures, by condition (ii) of Theorem 4.3 of Mendoza-Palacios and Hernández-Lerma (2015), there exists a constant $D >0$ such that

$$\begin{aligned} |\sigma _i(a_i|\mu (t))- \sigma _i(a_i|\mu (s))| \le D \; \Vert \mu (t)-\mu (s)\Vert _{\infty }. \end{aligned}$$

(23)

Using the replicator dynamics equation (2) and the boundedness of $\sigma _i$ (by condition (i) of Theorem 4.3 of Mendoza-Palacios and Hernández-Lerma (2015)) on $\bigtriangleup$, we get for $B_i \in {\mathcal {B}}(A_i)$,

$$\begin{aligned} |\mu _i(t)(B_i) - \mu _i(s)(B_i)|&= \left| \int _{s}^t \mu _i'(\zeta )(B_i) \; d\zeta \right| \nonumber \\&= \left| \int _{s}^t \int _{B_i} \sigma _i(a_i|\mu (\zeta )) \; \mu _i(\zeta )(da_i)\; d\zeta \right| \nonumber \\&\le C \; |t-s| \quad (\because \sigma _i \; \text {is bounded for all} \; i \in I). \end{aligned}$$

(24)

By (24), we have

$$\begin{aligned} \Vert \mu (t)-\mu (s)\Vert _{\infty } \le 2 \; C \; |t-s|. \end{aligned}$$

(25)

Substituting (25) in (23) gives

$$\begin{aligned} |\sigma _i(a_i|\mu (t))- \sigma _i(a_i|\mu (s))|&\le 2 \; C \; D \;|t-s|. \end{aligned}$$

This shows that the map $t \mapsto \sigma _i(a_i|\mu (t))$ is Lipschitz and hence uniformly continuous. $\square$

Lemma 5

Under the conditions and notations of part (b) in Theorem 1, we have

$$\begin{aligned} J_i(\eta _i(\cdot |{\bar{B}}), \eta _{-i}) - J_i(\eta _i,\eta _{-i}) = 0 \end{aligned}$$

for any weak cluster point $\eta$ of a replicator trajectory $(\mu (t))_{ t \ge 0}$.

Proof

Let $\eta$ be a weak cluster point of a trajectory $(\mu (t))_{t \ge 0}$. Recall that for $i \in I$, $\sigma _i(a_i|\eta ) = J_i(a_i, \eta _{-i}) - J_i(\eta _i, \eta _{-i})$. The set $G_i = \{a_i \in A_i \; | \; \sigma _i(a_i|\eta ) \ne 0 \}$ is open as the payoff functions $U_i$ are continuous.

From (20), it now follows that

$$\begin{aligned} \sigma _i(a_i|\eta ) = \lim \limits _{t \rightarrow \infty } \sigma _i(a_i|\mu (t)) = 0 \end{aligned}$$

for $\mu _i(0)$-almost every $a_i \in {\bar{B}}_i$, from which we obtain $\mu _i(0)(G_i) = 0$.

This immediately shows that

$$\begin{aligned} \mu _i(t)(G_i) = 0 \end{aligned}$$

(26)

because, by Lemma 2 of Bomze (1991) or Theorem 4.6 of Mendoza-Palacios and Hernández-Lerma (2015), the supports of $\mu _i(0)$ and $\mu _i(t)$ are same.

As $G_i$ is open and $\mu _i(t) \rightarrow \eta _i$ weakly, we get $\eta _i(G_i) \le \liminf _{t \rightarrow \infty } \mu _i(t)(G_i)$ by Theorem 2.1 (iv) of Billingsley (1999). Combining this fact with (26) yield $\eta _i(G_i) = 0$ which in turn gives

$$\begin{aligned} \eta _i(G_i \cap {\bar{B}}_i) = 0. \end{aligned}$$

(27)

Finally observe that

$$\begin{aligned} J_i(\eta _i(\cdot |{\bar{B}}), \eta _{-i}) - J_i(\eta _i,\eta _{-i})&= \frac{1}{\eta _i({\bar{B}}_i)} \int _{{\bar{B}}_i} \sigma _i(a_i|\eta ) \; \eta _i(da_i)\\&= \frac{1}{\eta _i({\bar{B}}_i)} \int _{G_i \cap {\bar{B}}_i} \sigma _i(a_i|\eta ) \; \eta _i(da_i) \quad (\because \sigma _i(a_i|\eta ) = 0 \; \text {on} \; G_i^c)\\&= 0 \quad (\text {by} \; (27) ). \end{aligned}$$

$\square$

Lemma 6

Assume that the pure strategy sets $A_1,\ldots ,A_n$ are compact. The norm $\Vert \cdot \Vert _\infty$ is weakly lower semicontinuous on $\bigtriangleup$. That is,

$$\begin{aligned} \Vert \mu \Vert _\infty \le \liminf _{k\rightarrow \infty }\Vert \mu ^{(k)}\Vert _\infty \end{aligned}$$

(28)

whenever $\mu ^{(k)}\rightarrow \mu$ weakly.

Proof

Let $\mu ^{(k)}\rightarrow \mu$ weakly. For every $i\in I$ and continuous function $f_i:A_i \rightarrow {\mathbb {R}}$ whose absolute value is bounded above by unity, we have

$$\begin{aligned} \left| \int _{A_i}f_i~d\mu _i^{(k)} \right|&\le \Vert \mu ^{(k)}_i \Vert \; (\text {by (1) in Mendoza-Palacios and Hern}\acute{\hbox {a}}\text {ndez-Lerma (2015)}\\&\quad \text {or p. 360 of Shiryaev (1995))}\\&\le \Vert \mu ^{(k)}\Vert _\infty ,~~k=1,2,\ldots . \end{aligned}$$

Taking limit infimum as $k\rightarrow \infty$ and using weak convergence of $\mu ^{(k)}$, we get

$$\begin{aligned} \left| \int _{A_i}f_i~d\mu _i \right| \le \liminf _{k \rightarrow \infty } \Vert \mu ^{(k)}\Vert _\infty . \end{aligned}$$

Now take supremum over all continuous $f_i$ (whose absolute value is bounded above by 1) and then maximum over $i\in I$, we get the desired inequality (28). $\square$

Lemma 7

Let $A_i$ be compact and ${{\bar{B}}}_i$ a closed subset of $A_i$, $i\in I$. Then the following hold true:

• (1) The face $\bigtriangleup _{{{\bar{B}}}}$ is weakly closed.

• (2) For every $\varepsilon >0$, the set $K^\varepsilon := \{ \mu \in \bigtriangleup ~:~ d_\infty (\mu ,\bigtriangleup _{{{\bar{B}}}}) \le \varepsilon \}$ is weakly closed.

Proof

• (1) Let $\{ \nu ^{(k)} \}$ be a sequence in $\bigtriangleup _{{{\bar{B}}}}$, and let $\nu ^{(k)} \rightarrow \nu$ weakly. To verify that $\nu \in \bigtriangleup _{{{\bar{B}}}}$, it is enough to show that $\nu _i \in (\bigtriangleup _i)_{{{\bar{B}}}_i}$, $i\in I$. Since ${{\bar{B}}}_i$ is closed and $\nu ^{({k})}_i \rightarrow \nu _i$ weakly, it follows that

  $$\begin{aligned} 1= \limsup _{k\rightarrow \infty } \nu ^{(k)}_i({{\bar{B}}}_i) \le \nu _i(\bar{B}_i) \end{aligned}$$

  from which we get $\nu _i({{\bar{B}}}_i)=1$. This implies that $\nu _i \in (\bigtriangleup _i)_{{{\bar{B}}}_i}$.

• (2) Let $\{ \mu ^{(k)} \}$ be a sequence in $K^\varepsilon$ and $\mu ^{(k)}\rightarrow \mu$ weakly. Since $\mu ^{(k)} \in K^\varepsilon$, we have

  $$\begin{aligned} \Vert \mu ^{(k)}-\nu ^{(k)} \Vert _{\infty } \le \varepsilon + \frac{1}{k} \end{aligned}$$

  (29)

  for some $\nu ^{(k)}\in \bigtriangleup _{{{\bar{B}}} }$, $k=1,2, \ldots$.

  As $A_i$ is compact, so is $\bigtriangleup _i$ with weak topology. This together with part (1) yields a subsequence $\{ \nu ^{k_\ell } \}_{\ell =1}^\infty$ which converges weakly to a probability measure (say $\nu$) in $\bigtriangleup _{{{\bar{B}}}}$. Taking limit infimum in (29) along this subsequence, we obtain

  $$\begin{aligned} \liminf _{\ell \rightarrow \infty } \Vert \mu ^{(k_\ell )}- \nu ^{(k_\ell )}\Vert _\infty \le \varepsilon . \end{aligned}$$

  Now we use Lemma 6 to conclude that $\Vert \mu -\nu \Vert _\infty \le \varepsilon$ and so $d_\infty (\mu ,\bigtriangleup _{{{\bar{B}}}})\le \varepsilon$. This shows that $\mu \in K^\varepsilon$ and hence the weak closedness of $K^\varepsilon$. $\square$